A box of mass m = 51 kg is being pulled by a constant force F = 101 N at an angle q = 15 degrees. (A) If the kinetic friction coefficient was m = 0.19 between the box and the surface, what is the magnitude of the friction force? What is the direction of the friction force? (B) What is the magnitude of the total, or net, force acting on the box?

A force |F| = 30 N is applied to the object with mass m = 2 kg. The angle theta = 25 degree. The object starts from rest and moves by displacement d = 2  m.  The coefficient of kinetic friction between the object and the floor is 0.2.  In this problem use g = 10m/s^2. The weight of the object is  W = mg. Find the work done on the object by the force F.  i hace calucated F = 54.37 From the above question, find the magnitude of the normal force between the object and the floor. Answer in Newton.   I have calculated magnutide of normal force = 32.67   From the above question, Find the work done on the object by friction. Watch out for the sign. I calculated  work done by friction = -13.06 QUESTION: From the above figure, find the final kinetic energy of the object. Answer in Joule.

A 45.9-kg box is being pushed a distance of 6.54 m across the floor by a force P whose magnitude is 168 N. The force P is parallel to the displacement of the box. The coefficient of kinetic friction is 0.208. Determine the work done on the box by (a) the applied force, (b) the friction force, (c) the normal force, and (d) by the force of gravity. Be sure to include the proper plus or minus sign for the work done by each force.

When the velocity of an object with mass 0.57 kg changes from 1.16 m/s to 0.32 m/s by the force F, what is the work done on the object by the force F?